Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $n = \dfrac{2k + 12}{k + 9} \div \dfrac{k^2 + 14k + 48}{-6k - 48} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $n = \dfrac{2k + 12}{k + 9} \times \dfrac{-6k - 48}{k^2 + 14k + 48} $ First factor the quadratic. $n = \dfrac{2k + 12}{k + 9} \times \dfrac{-6k - 48}{(k + 6)(k + 8)} $ Then factor out any other terms. $n = \dfrac{2(k + 6)}{k + 9} \times \dfrac{-6(k + 8)}{(k + 6)(k + 8)} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac{ 2(k + 6) \times -6(k + 8) } { (k + 9) \times (k + 6)(k + 8) } $ $n = \dfrac{ -12(k + 6)(k + 8)}{ (k + 9)(k + 6)(k + 8)} $ Notice that $(k + 8)$ and $(k + 6)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac{ -12\cancel{(k + 6)}(k + 8)}{ (k + 9)\cancel{(k + 6)}(k + 8)} $ We are dividing by $k + 6$ , so $k + 6 \neq 0$ Therefore, $k \neq -6$ $n = \dfrac{ -12\cancel{(k + 6)}\cancel{(k + 8)}}{ (k + 9)\cancel{(k + 6)}\cancel{(k + 8)}} $ We are dividing by $k + 8$ , so $k + 8 \neq 0$ Therefore, $k \neq -8$ $n = \dfrac{-12}{k + 9} ; \space k \neq -6 ; \space k \neq -8 $